= Solution
Let increasing overlap $x$ mean that the <dielectric> moves farther into the capacitor. Its <capacitance> derivative is
$$
\frac{dC}{dx}=\frac{2\pi\epsilon_0(\epsilon-1)}{\ln(b/a)}.
$$
At fixed <electric potential difference> $V$, the <capacitor energy> rises by $dU_{\rm field}=V^2dC/2$, but the voltage source supplies electrical <work> $V\,dQ=V^2dC$. The difference is available as mechanical <work> $F_xdx$:
$$
F_xdx=V\,dQ-dU_{\rm field}=\frac12V^2dC.
$$
The <dielectric insertion force at fixed voltage> is therefore
$$
\boxed{F_x=\frac12V^2\frac{dC}{dx}
=\frac{\pi\epsilon_0(\epsilon-1)V^2}{\ln(b/a)}.}
$$
For the usual $\epsilon>1$, it pulls the dielectric inward, increasing $x$; its magnitude is constant within the long-overlap approximation. In the original longitudinal diagram this is to the left. The units are newtons, since $V^2dC/dx$ is energy per length. Differentiating the field energy alone with a minus sign at fixed $V$ would miss the source contribution; the correct <fixed-voltage capacitor energy balance> is essential.
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